Question: $f(x) = x+4$ $g(t) = t^{2}+2t+5-4(f(t))$ $h(t) = 5t^{2}-5(f(t))$ $ h(f(6)) = {?} $
First, let's solve for the value of the inner function, $f(6)$ . Then we'll know what to plug into the outer function. $f(6) = 6+4$ $f(6) = 10$ Now we know that $f(6) = 10$ . Let's solve for $h(f(6))$ , which is $h(10)$ $h(10) = 5(10^{2})-5(f(10))$ To solve for the value of $h$ , we need to solve for the value of $f(10)$ $f(10) = 10+4$ $f(10) = 14$ That means $h(10) = 5(10^{2})+(-5)(14)$ $h(10) = 430$